| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: string becomes slack |
| Difficulty | Challenging +1.2 This is a standard vertical circle problem requiring energy conservation and the slack string condition (T=0). Part (a) is straightforward geometry, and part (b) applies standard mechanics principles. The multi-step nature and need to combine conditions elevates it slightly above average, but it follows a well-established template for M3 vertical circle questions. |
| Spec | 3.02h Motion under gravity: vector form6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sin \alpha = \frac{1}{2} \therefore \alpha = 30°\) \(\angle BOA = 90 + \alpha = 120°\) | M1 A1 | |
| resolve \(\curvearrowright\): \(T - mg\cos\theta = \frac{mv^2}{r}\) | M1 | |
| at B, \(T = 0, \theta = 120° \therefore v^2 = \frac{1}{2}ga\) | A1 | |
| con. of ME: \(\frac{1}{2}m(u^2 - v^2) = mg \times \frac{2}{3}a\) | M1 | |
| \(\therefore v^2 = u^2 - 3ga\) | A1 | |
| combining, \(\frac{1}{2}ga = u^2 - 3ga \therefore u^2 = \frac{7}{2}ga\) | M1 A1 | (8) |
| $\sin \alpha = \frac{1}{2} \therefore \alpha = 30°$ $\angle BOA = 90 + \alpha = 120°$ | M1 A1 | |
| resolve $\curvearrowright$: $T - mg\cos\theta = \frac{mv^2}{r}$ | M1 | |
| at B, $T = 0, \theta = 120° \therefore v^2 = \frac{1}{2}ga$ | A1 | |
| con. of ME: $\frac{1}{2}m(u^2 - v^2) = mg \times \frac{2}{3}a$ | M1 | |
| $\therefore v^2 = u^2 - 3ga$ | A1 | |
| combining, $\frac{1}{2}ga = u^2 - 3ga \therefore u^2 = \frac{7}{2}ga$ | M1 A1 | (8) |3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ad523c3f-9109-45a8-8399-80a4c2edeff7-3_513_570_196_625}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
A particle of mass $m$ is suspended at a point $A$ vertically below a fixed point $O$ by a light inextensible string of length $a$ as shown in Figure 2. The particle is given a horizontal velocity $u$ and subsequently moves along a circular arc until it reaches the point $B$ where the string becomes slack.
Given that the point $B$ is at a height $\frac { 1 } { 2 } a$ above the level of $O$,
\begin{enumerate}[label=(\alph*)]
\item show that $\angle B O A = 120 ^ { \circ }$,
\item show that $u ^ { 2 } = \frac { 7 } { 2 } g a$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q3 [8]}}